hdoj2602

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

Input

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

Output

One integer per line representing the maximum of the total value (this number will be less than 2³¹).

Sample Input

1

5 10

1 2 3 4 5

5 4 3 2 1

Sample Output

14

/*  * File:   main.cpp * Author: LeeningLu * * Created on 2012年3月27日, 下午3:01 */#include<stdio.h>#include<memory.h>#define M 1010int T;int N;//骨头数量int V;//背包容量struct bone{    int value;//骨头价值    int volume;//骨头所占背包容量}bone[M];int bag[M];int main(void){    scanf("%d",&T);    while(T--){        scanf("%d%d",&N,&V);//读取骨头数量和背包容量        for(int i=1;i<=N;i++)            scanf("%d",&bone[i].value);//读取每个骨头的价值        for(int i=1;i<=N;i++)            scanf("%d",&bone[i].volume);//读取每个骨头所占背包容量        memset(bag,0,sizeof(bag));//初始化数组        for(int j=1;j<=N;j++){            for(int i=V;i>=0;i--){//背包从大到小开始搜索是因为，如果从小到大开始搜索，有时会把一块骨头重复搜索两次                if(i>=bone[j].volume&&bone[j].value+bag[i-bone[j].volume]>bag[i])                    //状态转移方程max{bone[j].value+bag[i-bone[j].volume]，bag[i]}            }        }        printf("%d\n",bag[V]);    }    return 0;}